To find confidence intervals for the proportion of students who own a personal computer, we can use the formula for the confidence interval of a proportion:
\[ \hat{p} \pm z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]
where:
- \(\hat{p}\) is the sample proportion
- \(z\) is the z-score corresponding to the desired confidence level
- \(n\) is the sample size
Given:
- Sample size \(n = 500\)
- Number of students who own a personal computer = 420
- Sample proportion \(\hat{p} = \frac{420}{500} = 0.84\)
7.1 Find the 99% confidence interval
For a 99% confidence level, the z-score (\(z\)) is approximately 2.576.
Calculating the standard error:
\[ SE = \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} = \sqrt{\frac{0.84 \times (1 - 0.84)}{500}} = \sqrt{\frac{0.84 \times 0.16}{500}} = \sqrt{\frac{0.1344}{500}} = \sqrt{0.0002688} \approx 0.0164 \]
Now, calculating the confidence interval:
\[ \text{Margin of Error} = z \times SE = 2.576 \times 0.0164 \approx 0.0423 \] \[ \text{99% CI} = 0.84 \pm 0.0423 = (0.7977, 0.8823) \]
7.2 Find the 95% confidence interval
For a 95% confidence level, the z-score (\(z\)) is approximately 1.96.
Calculating the standard error (same as above, since it depends on \(n\) and \(\hat{p}\)):
Using \(SE \approx 0.0164\) (as previously calculated):
\[ \text{Margin of Error} = z \times SE = 1.96 \times 0.0164 \approx 0.0322 \] \[ \text{95% CI} = 0.84 \pm 0.0322 = (0.8078, 0.8722) \]
7.3 Find the 90% confidence interval
For a 90% confidence level, the z-score (\(z\)) is approximately 1.645.
Using \(SE \approx 0.0164\):
\[ \text{Margin of Error} = z \times SE = 1.645 \times 0.0164 \approx 0.0269 \] \[ \text{90% CI} = 0.84 \pm 0.0269 = (0.8131, 0.8669) \]
Summary of Confidence Intervals:
- 99% Confidence Interval: \((0.7977, 0.8823)\)
- 95% Confidence Interval: \((0.8078, 0.8722)\)
- 90% Confidence Interval: \((0.8131, 0.8669)\)